Ravi and Grace know there is a $1/3$ chance that two random players will tie each other in a given attempt at the game rock-paper-scissors. They plan to play $6$ attempts against each other. Assume that the results of subsequent attempts are independent, and let $T$ represent the number of attempts where they tie. Which of the following would find $P(T=2)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A ${6 \choose 2} \left( \dfrac13 \right)^2 \left( \dfrac23 \right)^4$ (Choice B) B ${6 \choose 2} \left( \dfrac13 \right)^4 \left( \dfrac23 \right)^2$ (Choice C) C ${6 \choose 3} \left( \dfrac13 \right)^2 \left( \dfrac23 \right)^4$ (Choice D) D ${6 \choose 3} \left( \dfrac13 \right)^4 \left( \dfrac23 \right)^2$ (Choice E) E $\left( \dfrac13 \right)^2 \left( \dfrac23 \right)^4$
Answer: Probability of $2$ ties We want the probability that there are $2$ successes (ties) in $6$ trials (attempts), so we're going to need $4$ failures (not ties) as well. The probability of each success is ${\dfrac13}$ and the probability of each failure is $\dfrac23}$. Since we're assuming independence, we can multiply probabilities to find the probability of getting $2$ successes followed by $4$ failures: $\begin{aligned} P(\text{SSFFFF})&=\left({\dfrac13}\right)\left({\dfrac13}\right)\left(\dfrac23}\right)\left(\dfrac23}\right)\left(\dfrac23}\right)\left(\dfrac23}\right) \\\\ &=\left({\dfrac13}\right)^2\left(\dfrac23}\right)^4 \end{aligned}$ The binomial coefficient ${n \choose k}$ SSFFFF isn't the only arrangement that produces $2$ successes in $6$ trials. For instance, FFFFSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=2$ successes (ties) in $n=6$ trials (attempts), so we should use the binomial coefficient ${6 \choose 2}$. [Tell me more about the binomial coefficient.] Putting it together Each arrangement has probability $\left( \dfrac13 \right)^2\left( \dfrac23 \right)^4$ so for our final answer we multiply this probability by the number of possible arrangements: ${6 \choose 2} \left( \dfrac13 \right)^2\left( \dfrac23 \right)^4$ The answer: ${6 \choose 2} \left( \dfrac13 \right)^2\left( \dfrac23 \right)^4$